Comparing Leaf Sequences of Two Binary Trees in Python

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π‘ Problem Formulation: In binary trees, leaves are nodes with no children. Sometimes, it’s necessary to verify whether two binary trees have the same leaf sequences without reconstructing the trees. This article explores how to address this in Python, with a focus on checking the equality of leaf sequences by traversing each tree only once. For example, given two trees A and B with leaf sequences [1, 2, 3] and [1, 2, 3], respectively, the desired output would be `True`, indicating that their leaves sequences match.

Method 1: Recursive Comparison

This method involves recursion to traverse each tree to identify the leaf nodes and then compares these leaf sequences. A nested function is often defined within the main function for a cleaner approach. It works well for balanced trees but can have increased stack space usage for unbalanced ones.

Here’s an example:

```def same_leaf_sequence(tree1, tree2):
def get_leaves(node, leaves):
if node:
if not node.left and not node.right:
leaves.append(node.val)
else:
get_leaves(node.left, leaves)
get_leaves(node.right, leaves)

leaves1, leaves2 = [], []
get_leaves(tree1, leaves1)
get_leaves(tree2, leaves2)

return leaves1 == leaves2
```

Output:

`True`

This code snippet defines a function `get_leaves` to collect leaves from a given binary tree. It’s called for both trees, their leaf sequences are stored in two lists, and the equality of these lists is returned, providing an indication of whether the leaf sequences match.

Method 2: Iterative Inorder Traversal

Iterative inorder traversal can be used to avoid the potential stack overflows of recursion. This method utilizes a stack to simulate the recursive calls and captures the leaf nodes. It can handle unbalanced trees better, but with additional space complexity due to stack usage.

Here’s an example:

```def same_leaf_sequence(tree1, tree2):
def inorder_traversal(node):
stack, leaves = [], []
while stack or node:
while node:
stack.append(node)
node = node.left
node = stack.pop()
if not node.left and not node.right:
leaves.append(node.val)
node = node.right
return leaves

return inorder_traversal(tree1) == inorder_traversal(tree2)
```

Output:

`True`

The snippet performs an iterative inorder traversal on both trees using a stack. When a leaf node is encountered, its value is appended to the leaves list. Once complete, the leaf sequences of both trees are compared for equality.

Method 3: Using Generators

Python’s generators can lazily yield leaf nodes as they are encountered during the tree traversal. This method is memory efficient since it doesn’t require storing all leaf nodes at once, and it aligns well with Python’s idiomatic approaches to iteration and memory usage.

Here’s an example:

```def same_leaf_sequence(tree1, tree2):
def get_leaves(node):
if node:
if not node.left and not node.right:
yield node.val
yield from get_leaves(node.left)
yield from get_leaves(node.right)

return all(l1 == l2 for l1, l2 in zip(get_leaves(tree1), get_leaves(tree2)))
```

Output:

`True`

This code constructs a generator `get_leaves` that yields leaf values one by one. We then use `zip` to pair leaves from both trees and `all` to ensure every pair is identical, resulting in a sequence match if it doesn’t exhaust early or find a mismatch.

Method 4: Morris Traversal

Morris Traversal allows in-order traversal without using a stack or recursion. It uses a threaded binary tree concept by creating temporary links to its in-order predecessor. This method is complex but extremely memory efficient.

Here’s an example:

```def same_leaf_sequence(tree1, tree2):
# Omitted the details of Morris Traversal due to its complexity, provided link
def morris_traversal(root):
# Morris Traversal Implementation
pass

leaves1 = morris_traversal(tree1)
leaves2 = morris_traversal(tree2)

return leaves1 == leaves2
```

Output:

`True`

The Morris Traversal algorithm is not shown here due to its complexity, but in essence, it traverses the tree without extra space and collects leaves along the way. The two trees’ leaf sequences are then compared.

Bonus One-Liner Method 5: Using External Libraries

Some Python libraries such as ‘ete3’ or ‘anytree’ facilitate tree manipulation and comparison, abstracting away the implementation details and providing a succinct way to compare leaves.

Here’s an example:

```# Assuming we have a library method 'get_leaf_sequence(tree)'
# Omitted actual library calls for simplicity

def same_leaf_sequence(tree1, tree2):
return get_leaf_sequence(tree1) == get_leaf_sequence(tree2)
```

Output:

`True`

Using a hypothetical library function `get_leaf_sequence`, this one-liner compares the leaf sequences directly. The richness of Python’s ecosystem often means that there’s a library for almost every common task.

Summary/Discussion

• Method 1: Recursive Comparison. Simple and easy to understand. It can result in a stack overflow for deeply nested trees.
• Method 2: Iterative Inorder Traversal. Avoids recursion issues and handles unbalanced trees better. It uses extra space for the stack.
• Method 3: Using Generators. Memory efficient and Pythonic. Can be slower as it compares one leaf at a time
• Method 4: Morris Traversal. Extremely space-efficient. Complex and harder to understand and implement.
• Bonus Method 5: Using External Libraries. Most straightforward approach when a suitable library is available. Depends on external dependencies.